18. SOLUBILITY AND COMPLEX-ION EQUILIBRIAteachers.sduhsd.net/kshakeri/oldsite/ap_chemistry_pages/...SOLUBILITY AND COMPLEX-ION EQUILIBRIA Solutions to Exercises Note on significant

18. SOLUBILITY AND COMPLEX-ION EQUILIBRIA

Solutions to Exercises Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off.

18.1 a. BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+][SO4

2-] b. Fe(OH)3(s) Fe3+(aq) + 3OH-(aq) Ksp = [Fe3+][OH-]3 c. Ca3(PO4)2(s) 3Ca2+(aq) + 2PO4

3-(aq); Ksp = [Ca2+]3[PO43-]2

18.2 Calculate the molar solubility. Then, assemble the usual concentration table, and

substitute the equilibrium concentrations from it into the equilibrium-constant expression. (Because no concentrations can be given for solid AgCl, dashes are written; in later problems, similar spaces will be left blank.)

-31.9 x 10 g

1 L x 1 mol

143 g = 1.33 x 10-5 M

(continued)

SOLUBILITY AND COMPLEX-ION EQUILIBRIA 679

Set up the table as usual.

Conc. (M) AgCl(s) Ag+ + Cl-

Starting 0 0

Change +1.33 x 10-5 +1.33 x 10-5

Equilibrium 1.33 x 10-5 1.33 x 10-5 Ksp = [Ag+][Cl-] = (1.33 x 10-5)(1.33 x 10-5) = 1.768 x 10-10 = 1.8 x 10-10 18.3 Calculate the molar solubility. Then, assemble the usual concentration table and

substitute from it into the equilibrium-constant expression. (Because no concentrations can be given for solid Pb3(AsO4)2, spaces are left blank.)

-53.0 x 10 g

1 L x 1 mol

899 g = 3.34 x 10-8 M

Conc. (M) Pb3(AsO4)2(s) 3Pb2+ + 2AsO4

3-

Starting 0 0

Change +3(3.34 x 10-8) +2(3.34 x 10-8)

Equilibrium 3(3.34 x 10-8) 2(3.34 x 10-8) Ksp = [Pb2+]3[AsO4

3-]2 = [3 x (3.34 x 10-8)]3(2 x 3.34 x 10-8)2

= 4.489 x 10-36 = 4.5 x 10-36 18.4 Assemble the usual concentration table. Let x equal the molar solubility of CaSO4.

When x mol CaSO4 dissolves in one L of solution, x mol Ca2+ and x mol SO42- form.

Conc. (M) CaSO4(s) Ca2+ + SO4

2-

Starting 0 0

Change +x +x

Equilibrium x x Substitute the equilibrium concentrations into the equilibrium-constant expression and

solve for x. Then, convert to g CaSO4 per L.

(continued)

680 CHAPTER 18

[Ca2+][SO42-] = Ksp

(x)(x) = x2 = 2.4 x 10-5

x = )10 x (2.4 5- = 4.89 x 10-3 M

L

mol 10 x 4.89 -3 x 136 g

1 mol

= 0.6664 = 0.67 g/L

18.5 a. Let x equal the molar solubility of BaF2. Assemble the usual concentration table, and

substitute from the table into the equilibrium-constant expression.

Conc. (M) BaF2(s) Ba2+ + 2F-

Starting 0 0

Change +x +2x

Equilibrium x 2x [Ba2+][F-]2 = Ksp (x)(2x)2 = 4x3 = 1.0 x 10-6

x = 36-

410 x 1.0 = 6.299 x 10-3 = 6.3 x 10-3 M

b. At the start, before any BaF2 dissolves, the solution contains 0.15 M F-. At equilibrium,

x mol of solid BaF2 dissolves to yield x mol Ba2+ and 2x mol F-. Assemble the usual concentration table, and substitute the equilibrium concentrations into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.15 M F-.

Conc. (M) BaF2(s) Ba2+ + 2F-

Starting 0 0.15

Change +x +2x

Equilibrium x 0.15 + 2x [Ba2+][F-]2 = Ksp

(continued)


(x)(0.15 + 2x)2 ≅ (x)(0.15)2 ≅ 1.0 x 10-6

x ≅ 2

-6

(0.15)10 x 1.0 = 4.444 x 10-5 = 4.4 x 10-5 M

Note that adding 2x to 0.15 M will not change it (to two significant figures), so 2x is

negligible compared to 0.15 M. The solubility of 4.4 x 10-5 M in 0.15 M NaF is lower than the solubility of 6.3 x 10-3 M in pure water.

18.6 Calculate the ion product, Qc, after evaporation, assuming no precipitation has

occurred. Compare it with the Ksp. Qc = [Ca2+][SO4

2-] Qc = (2 x 0.0052)(2 x 0.0041) = 8.528 x 10-5 Since Qc > Ksp (2.4 x 10-5), precipitation occurs. 18.7 Calculate the concentrations of Pb2+ and SO4

2-, assuming no precipitation. Use a total volume of 0.456 L + 0.255 L, or 0.711 L.

[Pb2+] =

0.00016 mol x 0.255 LL

0.711 L = 5.74 x 10-5 M

[SO42-] =

0.00023 mol x 0.456 LL

0.711 L = 1.48 x 10-4 M

Calculate the ion product, and compare it to Ksp.

Qc = [Pb2+][SO4

2-] = (5.74 x 10-5)(1.48 x 10-4) = 8.49 x 10-9 Because Qc is less than the Ksp of 1.7 x 10-8, no precipitation occurs, and the solution

is unsaturated.

682 CHAPTER 18

18.8 The solubility of AgCN would increase as the pH decreases because the increasing concentration of H3O+ would react with the CN- to form the weakly ionized acid HCN. As CN- is removed, more AgCN dissolves to replace the cyanide:

AgCN(s) Ag+(aq) + CN-(aq) [+ H3O+ → HCN + H2O] In the case of AgCl, the chloride ion is the conjugate base of a strong acid and would,

therefore, not be affected by any amount of hydrogen ion. 18.9 Because Kf = 4.8 x 1012 and because the starting concentration of NH3 is much larger

than that of the Cu2+ ion, you can make a rough assumption that most of the copper(II) is converted to Cu(NH3)4

2+ ion. This ion then dissociates slightly to give a small concentration of Cu2+ and additional NH3. The amount of NH3 remaining at the start after reacting with 0.015 M Cu2+ is

[0.100 M - (4 x 0.015 M)] = 0.040 M starting NH3

Assemble the usual concentration table using this starting concentration for NH3 and

assuming the starting concentration of Cu2+ is zero.

Conc. (M) Cu(NH3)42+ Cu2+ + 4NH3

Starting 0.015 0 0.040

Change -x +x +4x

Equilibrium 0.015 - x x 0.040 + 4x Even though this reaction is the opposite of the equation for the formation constant,

the formation-constant expression can be used. Simply substitute all exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.015 and 4x is negligible compared to 0.040.

Kf = 2+

3 42+ 4

3

u(NH ) ][Cu ][NH ][C = 4

(0.015 - x)(x)(0.040 + 4x)

≅ 4(0.015)

(x)(0.040) ≅ 4.8 x 10-12

Rearrange and solve for x:

x = [Cu2+] ≅ (0.015) ÷ [(4.8 x 1012)(0.040)4] ≅ 1.22 x 10-9 = 1.2 x 10-9 M


18.10 Start by calculating the [Ag+] in equilibrium with the Ag(CN)2- formed from Ag+ and CN-.

Then, use the [Ag+] to decide whether or not AgI will precipitate by calculating the ion product and comparing it with the Ksp of 8.3 x 10-17 for AgI. Assume all the 0.0045 M Ag+ reacts with CN- to form 0.0045 M Ag(CN)2

-, and calculate the remaining CN-. Use these as starting concentrations for the usual concentration table.

[0.20 M KCN - (2 x 0.0045 M)] = 0.191 M starting CN-

Conc. (M) Ag(CN)2

- Ag+ + 2CN-

Starting 0.0045 0 0.191

Change -x +x +2x


the formation-constant expression can be used. Simply substitute all exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.0045 and 2x is negligible compared to 0.191.

Kf = -

2+ -

Ag(CN) ][Ag ][CN ]2[ = 2

(0.0045 - x)(x)(0.191 + 2x)

≅ 2(0.0045)

(x)(0.191) ≅ 5.6 x 1018


x = [Ag+] ≅ (0.0045) ÷ [(5.6 x 1018)(0.191)2] ≅ 2.202 x 10-20 M Now, calculate the ion product for AgI: Qc = [Ag+][I-] = (2.20 x 10-20)(0.15) = 3.30 x 10-21 = 3.3 x 10-21 Because Qc is less than the Ksp of 8.3 x 10-17, no precipitate will form, and the solution

is unsaturated.

684 CHAPTER 18

18.11 Obtain the overall equilibrium constant for this reaction from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation:

AgBr(s) Ag+(aq) + Br-(aq) Ksp = 5.0 x 10-13

Ag+(aq) + 2S2O32-(aq) Ag(S2O3)2

3-(aq) Kf = 2.9 x 1013

AgBr(s) + 2S2O32-(aq) Ag(S2O3)2

3-(aq) + Br-(aq) Kc = Ksp x Kf = 14.5

Assemble the usual table using 1.0 M as the starting concentration of S2O32- and x as

the unknown concentration of Ag(S2O3)23- formed.

Conc. (M) AgBr(s) + 2S2O3

2- Ag(S2O3)23- + Br-

Starting 1.0 0 0

Change -2x +x +x

Equilibrium 1.0 - 2x x x The equilibrium-constant expression can now be used. Simply substitute all exact

equilibrium concentrations into the equilibrium-constant expression. The solution can be obtained without using the quadratic equation.

Kc = 3- -

2 3 22- 2

2 3

g(S O ) ][Br ]

[S O ]

[A = 2

2(x)

(1.0 - 2x) = 14.5

Take the square root of both sides of the two right-hand terms, and solve for x:

0883. = )x2 - 1.0(

x

x = 3.808 (1.0 - 2x) 7.62x + x = 3.808

x = 0.4417 = 0.44 M (Molar solubility of AgBr in 1.0 M Na2S2O3)


Answers to Concept Checks

18.1 Solubility and Ksp are related, although not directly. You can compare Ksp's for a series of salts, however, if they have the same number of cations and anions in each of their formulas. (In that case, Ksp and solubility are related in the same way for each salt.) In this problem, each of the lead(II) compounds has one Pb2+ cation and one anion, so you can compare the Ksp's directly. Lead(II) sulfate has the largest Ksp and, therefore, is the most soluble of these lead(II) compounds.

18.2 Let's look at each compound in turn. NaNO3 has no ion in common with PbSO4, so it

should have little effect on its solubility. Na2SO4 is a soluble compound and provides the common ion SO4

2-, which would repress the solubility of PbSO4. PbS has an ion in common with PbSO4 (Pb2+), but the compound is so insoluble that very little of the Pb2+ ion is available. Because of this, the solubility of PbSO4 is little affected by the PbS. Therefore, the NaNO3 solution will dissolve the most PbSO4.

18.3 If NaCl were added to a saturated AgCl solution, the equilibrium would shift to

consume the added chloride ion, and some AgCl would precipitate. After the addition of two Cl- ions, there would be five Cl- ions, three Ag+ ions, and two AgCl molecules. The solution would look like the following (Na+ not shown for clarity).

AgCl

Cl-

Ag+

18.4 If you compare Ksp’s for magnesium oxalate (8.5 x 10-5) and calcium oxalate (2.3 x 10-

9), you can see magnesium oxalate is much more soluble in water solution than the calcium salt (the Ksp is larger). This means it provides a greater concentration of oxalate ion. In water solution, some of the magnesium oxalate dissolves, giving an oxalate ion concentration that tends to repress the dissolution of calcium oxalate (common-ion effect). The addition of acid tends to remove oxalate ion, but this is replenished by the dissolution of more magnesium oxalate. Therefore, you would expect the magnesium oxalate to be more likely to dissolve.

686 CHAPTER 18

Answers to Review Questions

18.1 The solubility equation is Ni(OH)2(s) Ni2+(aq) + 2OH-(aq). If the molar solubility of Ni(OH)2 = x molar, the concentrations of the ions in the solution must be x M Ni2+ and 2x M OH-. Substituting into the equilibrium-constant expression gives

Ksp = [Ni2+][OH-]2 = (x)(2x)2 = 4x3 18.2 Calcium sulfate is less soluble in a solution containing sodium sulfate because the

increase in sulfate from the sodium sulfate causes the equilibrium composition in the equation below to shift to the left:

CaSO4(s) Ca2+(aq) + SO4

2-(aq) The result is a decrease in both the calcium ion and the calcium sulfate

concentrations. 18.3 Substitute the 0.10 M concentration of chloride into the solubility product expression

and solve for [A+]:

[Ag+] = sp-

K

[Cl ] =

-101.8 x 100.10

= 1.80 x 10-9 = 1.8 x 10-9 M

18.4 In order to predict whether or not PbI2 will precipitate when lead nitrate and potassium

iodide are mixed, the concentrations of Pb2+ and I- after mixing would have to be calculated first (if the concentrations are not known or are not given). Then, the value of Qc, the ion product, would have to be calculated for PbI2. Finally, Qc would have to be compared with the value of Ksp. If Qc > Ksp, then a precipitate will form at equilibrium. If Qc is ≤ than Ksp, no precipitate will form.

18.5 Barium fluoride, normally insoluble in water, dissolves in dilute hydrochloric acid

because the fluoride ion, once it forms, reacts with the hydronium ion to form weakly ionized HF:

BaF2(s) Ba2+(aq) + 2 F-(aq) [+ 2 H3O+ → 2 HF + 2 H2O] 18.6 Metal ions such as Pb2+ and Zn2+ are separated by controlling the [S2-] in a solution of

saturated H2S by means of adjusting the pH correctly. Because the Ksp of 2.5 x 10-27 for PbS is smaller than the Ksp of 1.1 x 10-21 for ZnS, the pH can be adjusted to make the [S2-] just high enough to precipitate PbS without precipitating ZnS.


18.7 When NaCl is first added to a solution of Pb(NO3)2, a precipitate of PbCl2 forms. As more NaCl is added, the excess chloride reacts further with the insoluble PbCl2, forming soluble complex ions of PbCl3- and PbCl42-:

Pb2+(aq) + 2Cl-(aq) PbCl2(s) PbCl2(s) + Cl-(aq) PbCl3-(aq) PbCl3-(aq) + Cl-(aq) PbCl42-(aq) 18.8 When a small amount of NaOH is added to a solution of Al2(SO4)3, a precipitate of

Al(OH)3 forms at first. As more NaOH is added, the excess hydroxide ion reacts further with the insoluble Al(OH)3, forming a soluble complex ion of Al(OH)4

-. 18.9 The Ag+, Cu2+, and Ni2+ ions can be separated in two steps: (1) Add HCl to precipitate

just the Ag+ as AgCl, leaving the others in solution. (2) After pouring the solution away from the precipitate, add 0.3 M HCl and H2S to precipitate only the CuS away from the Ni2+ ion, whose sulfide is soluble under these conditions.

18.10 By controlling the pH through the appropriate buffer, one can control the [CO3

2-] using the equilibrium reaction:

H3O+(aq) + CO3

2-(aq) HCO3-(aq) + H2O(l)

Calcium carbonate is much more insoluble than magnesium carbonate and thus will

precipitate in weakly basic solution, whereas magnesium carbonate will not. Magnesium carbonate will precipitate only in highly basic solution. (There is the possibility that Mg(OH)2 might precipitate, but the [OH-] is too low for this to occur.)

Answers to Conceptual Problems

18.11 a. Since both compounds contain the same number of ions, the salt with the larger Ksp value will be more soluble. The Ksp for AgCl is 1.8 x 10-10 and for AgI is 8.3 x 10-17. Therefore, silver chloride (AgCl) is more soluble.

b. Since both compounds contain the same number of ions, the salt with the larger Ksp

value will be more soluble. The Ksp for Mg(OH)2 is 1.8 x 10-11 and for Cu(OH)2 is 2.6 x 10-19. Therefore, magnesium hydroxide (Mg(OH)2) is more soluble.

18.12 Dissolve a sample of each mineral in water in a separate container. NaCl is soluble in

water, and CaF2 is insoluble in water (Ksp = 3.4 x 10-11), so the sample that dissolves is NaCl, and the sample that doesn’t dissolve is CaF2.

18.13 0.1 M NaCl will be most effective in causing a precipitate from a saturated solution of

PbCl2, since it is completely soluble and contains the common ion Cl-. Therefore, the answer is a.

688 CHAPTER 18

18.14 The beaker on the left depicts all the AgCl as individual formula units in solution. This implies AgCl is a soluble nonelectrolye, which is not the case since AgCl as a slightly soluble ionic compound. The center beaker depicts AgCl as a soluble ionic compound that completely dissolves in solution, leaving no AgCl(s). Once again, this cannot be correct since AgCl is slightly soluble. The beaker on the right indicates there are ions of Ag+ and Cl- present in the solution along with solid AgCl. This is consistent with AgCl being a slightly soluble ionic compound.

18.15 The beaker on the left depicts all the NaCl as individual formula units in solution. This

implies NaCl is a soluble nonelectrolye, which is not the case since NaCl is a very soluble ionic compound that produces ions in solution. The center beaker depicts NaCl as a soluble ionic compound that completely dissolves in solution, producing as only Na+(aq) and Cl-(aq). This must be correct since NaCl is a very soluble ionic compound. The beaker on the right indicates there are ions of Na+ and Cl- present in the solution along with solid NaCl. The presence of solid is not consistent with NaCl being a very soluble ionic compound.

18.16 When first added to a solution of copper(II) nitrate, the ammonia causes a pale blue

precipitate of Cu(OH)2 to form (Ksp = 2.6 x 10-19). After enough ammonia has been added, however, the complex ion Cu(NH3)4

2+ forms (Kf = 4.8 x 1012), and the precipitate dissolves.

18.17 Add just enough Na2SO4 to precipitate all the Ba2+; filter off the BaSO4; add more

Na2SO4 to precipitate all the Ca2+; filter off the CaSO4; Mg2+ remains in the solution. 18.18 When a precipitate fails to form when HCl is added to the solution, this indicates no

silver ion is present. When a precipitate fails to form when the solution is acidified and H2S is added, this indicates no copper(II) ion is present in the solution.

Solutions to Practice Problems Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off.

18.19 a. NaBr is soluble (Group IA salts are soluble). b. PbI2 is insoluble (PbI2 is an insoluble iodide). c. BaCO3 is insoluble (Carbonates are generally insoluble). d. (NH4)2SO4 is soluble (All ammonium salts are soluble).


18.20 a. Ca(NO3)2 is soluble (Nitrate salts are soluble). b. AgBr is insoluble (AgBr is an insoluble bromide salt). c. MgI2 is soluble (Iodide salts are generally soluble). d. PbSO4 is insoluble (PbSO4 is an insoluble sulfate salt). 18.21 a. Ksp = [Mg2+][OH-]2 b. Ksp = [Sr2+][CO3

2-] c. Ksp = [Ca2+]3[AsO4

3-]2 d. Ksp = [Fe3+][OH-]3 18.22 a. Ksp = [Ba2+]3[PO4

3-]2 b. Ksp = [Fe3+][PO43-]

c. Ksp = [Pb2+][I-]2 d. Ksp = [Ag+]2[S2-] 18.23 Calculate molar solubility, assemble the usual table, and substitute the equilibrium

concentrations from it into the equilibrium-constant expression. (Dashes are given for AgBrO3(s); in later problems, the dashes are omitted.)

-37.2 x 10 g

L x 1 mol

236 g = 3.05 x 10-5 M

Conc. (M) AgBrO3(s) Ag+ + BrO3

-

Starting — 0 0

Change — +3.05 x 10-5 +3.05 x 10-5

Equilibrium — 3.05 x 10-5 3.05 x 10-5 Ksp = [Ag+][BrO3

-] = (3.05 x 10-5)(3.05 x 10-5) = 9.302 x 10-10 = 9.3 x 10-10 18.24 Assemble the usual concentration table, and substitute from it the equilibrium

concentrations into the equilibrium-constant expression.

Conc. (M) MgC2O4(s) Mg2+ + C2O42-

Starting 0 0

Change +9.3 x 10-3 +9.3 x 10-3

Equilibrium 9.3 x 10-3 9.3 x 10-3 Ksp = [Mg2+][C2O4

2-] = (9.3 x 10-3)(9.3 x 10-3) = 8.649 x 10-5 = 8.6 x 10-5

690 CHAPTER 18

18.25 Calculate the molar solubility. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression.

0.13 g0.100 L

x 1 mol413 g

= 3.147 x 10-3 M

Conc. (M) Cu(IO3)2(s) Cu2+ + 2IO3-

Starting 0 0

Change +3.147 x 10-3 +2 x 3.147 x 10-3

Equilibrium 3.147 x 10-3 2 x 3.147 x 10-3

Ksp = [Cu2+][IO3-]2 = (3.147 x 10-3)[2(3.147 x 10-3)]2 = 1.247 x 10-7 = 1.2 x 10-7

18.26 Calculate the molar solubility. Then, assemble the usual concentration table, and

substitute the equilibrium concentrations from it into the equilibrium-constant expression.

0.022 gL

x 1 mol332 g

= 6.62 x 10-5 M

Conc. (M) Ag2CrO4(s) 2Ag+ + CrO42-

Starting 0 0

Change +2(6.62 x 10-5) 6.62 x 10-5

Equilibrium 2(6.62 x 10-5) 6.62 x 10-5

Ksp = [Ag+]2[CrO42-] = [2(6.62 x 10-5)]2(6.62 x 10-5) = 1.16 x 10-12 = 1.2 x 10-12

18.27 Calculate the pOH from the pH, and then convert pOH to [OH-]. Then, assemble the

usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression.

pOH = 14.00 - 10.52 = 3.48

[OH-] = antilog (-pOH) = antilog (-3.48) = 3.311 x 10-4 M

[Mg2+] = [OH-] ÷ 2 = (3.311 x 10-4) ÷ 2 = 1.655 x 10-4 M

Conc. (M) Mg(OH)2(s) Mg2+ + 2OH-

Starting 0 0

Change +1.655 x 10-4 +3.311 x 10-4

Equilibrium 1.655 x 10-4 3.311 x 10-4

Ksp = [Mg2+][OH-]2 = (1.655 x 10-4)(3.311 x 10-4)2 = 1.81 x 10-11 = 1.8 x 10-11


18.28 Calculate the pOH from the pH, and then convert pOH to [OH-]. pOH = 14.00 - 12.35 = 1.65 [OH-] = antilog (-pOH) = antilog (-1.65) = 2.238 x 10-2 M [Ca2+] = [OH-] ÷ 2 = (2.238 x 10-2) ÷ 2 = 1.119 x 10-2 M Assemble the usual concentration table, and substitute the equilibrium concentrations

from it into the equilibrium-constant expression.

Conc. (M) Ca(OH)2(s) Ca2+ + 2OH-

Starting 0 0

Change +1.119 x 10-2 +2.238 x 10-2

Equilibrium 1.119 x 10-2 2.238 x 10-2 Ksp = [Ca2+][OH-]2 = (1.119 x 10-2)(2.238 x 10-2)2 = 5.60 x 10-6 = 5.6 x 10-6 18.29 Assemble the usual concentration table. Let x equal the molar solubility of SrCO3.

When x mol SrCO3 dissolves in one L of solution, x mol Sr2+ and x mol CO32- form.

Conc. (M) SrCO3(s) Sr2+ + CO3

2-

Starting 0 0

Change +x +x


solve for x. Then, convert to grams SrCO3 per liter. [Sr2+][CO3

2-] = Ksp

(x)(x) = x2 = 9.3 x 10-10

x = )10 x (9.3 10- = 3.04 x 10-5 M

L

mol 10 x 3.04 -5 x

3SrCO mol 1g 147.63 = 4.50 x 10-3 = 0.0045 g/L

692 CHAPTER 18

18.30 Assemble the usual concentration table. Let x equal the molar solubility of MgCO3. When x mol MgCO3 dissolves in one L of solution, x mol Mg2+ and x mol CO3

2- form.

Conc. (M) MgCO3(s) Mg2+ + CO32-

Starting 0 0

Change +x +x


solve for x. Then, convert to grams MgCO3 per liter. [Mg2+][CO3

2-] = Ksp

(x)(x) = x2 = 1.0 x 10-5

x = )10 x (1.0 5- = 3.162 x 10-3 M

L

mol 10 x 3.162 -3 x

3MgCO mol 1g 84.32 = 0.2666 = 0.27 g/L

18.31 Let x equal the molar solubility of PbF2. Assemble the usual concentration table, and

substitute from the table into the equilibrium-constant expression.

Conc. (M) PbF2(s) Pb2+ + 2F-

Starting 0 0

Change +x +2x

Equilibrium x 2x [Pb2+][F-]2 = Ksp (x)(2x)2 = 4x3 = 2.7 x 10-8

x = 38-

410 x 2.7 = 1.88 x 10-3 = 1.9 x 10-3 M


18.32 Let x equal the molar solubility of MgF2. Assemble the usual concentration table.

Conc. (M) MgF2(s) Mg2+ + 2F-

Starting 0 0

Change +x +2x

Equilibrium x 2x Substitute from the table into the equilibrium-constant expression [Mg2+][F-]2 = Ksp (x)(2x)2 = 4x3 = 7.4 x 10-11

x = 311-

410 x 7.4 = 2.644 x 10-4 = 2.6 x 10-4 M

18.33 At the start, before any SrSO4 dissolves, the solution contains 0.23 M SO4

2-. At equilibrium, x mol of solid SrSO4 dissolves to yield x mol Sr2+ and x mol SO4

2-. Assemble the usual concentration table, and substitute the equilibrium concentrations into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.23 M SO4

2-.

Conc. (M) SrSO4(s) Sr2+ + SO42-

Starting 0 0.23

Change +x +x

Equilibrium x 0.23 + x [Sr2+][SO4

2-] = Ksp

(x)(0.23 + x) ≅ (x)(0.23) ≅ 2.5 x 10-7

x ≅ -7 10

0.232.5 x = 1.086 x 10-6 M

-61.086 x 10 mol

L x

4

183.69 g1 mol SrSO = 1.99 x 10-4 = 2.0 x 10-4 g/L

Note that adding x to 0.23 M will not change it (to two significant figures), so x is

negligible compared to 0.23 M.

694 CHAPTER 18

18.34 Let x = the molar solubility of PbCrO4. At the start, before any PbCrO4 dissolves, the solution contains 0.13 M CrO4

2-. At equilibrium, x mol of solid PbCrO4 dissolves to yield x mol Pb2+ and x mol CrO4

2-. Assemble the usual concentration table, and substitute the equilibrium concentrations into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.13 M CrO4

2-.

Conc. (M) PbCrO4(s) Pb2+ + CrO42-

Starting 0 0.13

Change +x +x

Equilibrium x 0.13 + x [Pb2+][CrO4

2-] = Ksp

(x)(0.13 + x) ≅ (x)(0.13) ≅ 1.8 x 10-14

x ≅ -14 x 10

0.131.8 = 1.38 x 10-13 M

-131.38 x 10 mol

L x

4

323.2 g1 mol PbCrO =4.47 x 10-11 = 4.5 x 10-11 g/L

Note that adding x to 0.13 M will not change it (to two significant figures), so x is

negligible compared to 0.13 M. 18.35 Calculate the value of Ksp from the solubility using the concentration table. Then, using

the common-ion calculation, assemble another concentration table. Use 0.020 M NaF as the starting concentration of F- ion. Substitute the equilibrium concentrations from the table into the equilibrium-constant expression. As an approximation, assume 2x is negligible compared to 0.020 M F- ion.

L

g 0.016 x 21 mol MgF62.31 g

= 2.567 x 10-4 M


Starting 0 0

Change +2.567 x 10-4 +2 x 2.567 x 10-4

Equilibrium 2.567 x 10-4 2 x 2.567 x 10-4

(continued)


Ksp = [Mg2+][F-]2 = (2.567 x 10-4)[2(2.567 x 10-4)]2 = 6.772 x 10-11 Now, use Ksp to calculate the molar solubility of MgF2.


Starting 0 0.020

Change +x +2x

Equilibrium x 0.020 + 2x [Mg2+][F-]2 = Ksp

(x)(0.020 + 2x)2 ≅ (x)(0.020)2 = 6.772 x 10-11

x ≅ -11

22 x 10

(0.020)6.77 = 1.693 x 10-7 M

-71.693 x 10 mol

L x

2MgF mol 1g 62.31 = 1.054 x 10-5 = 1.1 x 10-5 g/L

18.36 Calculate the value of Ksp from the solubility using the concentration table. Then, using

the common-ion calculation, assemble another concentration table. Use 0.45 M Na2SO4 as the starting concentration of SO4

2- ion. Substitute the equilibrium concentrations from the table into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.45 M SO4

2- ion.

L

g 8.0 x 2 41 mol Ag SO311.81 g

= 0.02565 M

Conc. (M) Ag2SO4(s) 2Ag+ + SO4

2-

Starting 0 0

Change +2 x 0.02565 +0.02565

Equilibrium 2 x 0.02565 0.02565 Ksp = [Ag+]2[SO4

2-] = (2 x 0.02565)2(0.02565) = 6.755 x 10-5 Now, use Ksp to calculate the molar solubility of Ag2SO4.

(continued)

696 CHAPTER 18

Conc. (M) Ag2SO4(s) 2Ag+ + SO42-

Starting 0 0.45

Change +2x +x

Equilibrium 2x 0.45 + x [Ag+]2[SO4

2-] = Ksp (2x)2(0.45 + x) ≅ (2x)2(0.45) = 6.755 x 10-5

x ≅ -56.755 x 10 )

4 x 0.45( = 6.125 x 10-3 M

-36.125 x 10 mol

L x

2 4

311.81 g1 mol Ag SO = 1.910 = 1.9 g/L

18.37 The concentration table follows.

Conc. (M) MgC2O4(s) Mg2+ + C2O42-

Starting 0 0.020

Change +x +x

Equilibrium x 0.020 + x The equilibrium-constant equation is Ksp = [Mg2+][C2O4

2-]

8.5 x 10-5 = x(0.020 + x) x2 + 0.020x - (8.5 x 10-5) = 0 Solving the quadratic equation gives

x = 2

)10 x 4(8.5 (0.020) 0.020- 5-2 +± = 0.0036 M

The solubility in grams per liter is

L

mol 0.0036 x 112 g1 mol

= 0.403 = 0.4 g/L


18.38 The concentration table follows.

Conc. (M) SrSO4(s) Sr2+ + SO42-

Starting 0 0.0015

Change +x +x

Equilibrium x 0.0015 + x Ksp = 2.5 x 10-7 = x(0.0015 + x) x2 + 0.0015x - (2.5 x 10-7) = 0 Solving the quadratic equation gives

x = 2 --0.0015 (0.0015) + 4(2.5 x 10 )

2± 7

= 1.51 x 10-4 M

The solubility in grams per liter is

L

mol 10 x 1.51 -4 x

4

183.69 g1 mol SrSO = 0.0277 = 0.03 g/L

18.39 a. Calculate Qc, the ion product of the solution, using the concentrations in the problem

as the concentrations present after mixing and assuming no precipitation. Then, compare Qc with Ksp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of Ksp; then, use parentheses for the concentrations.

Qc = [Ba2+][F-]2 Qc = (0.020)(0.015)2 = 4.50 x 10-6 Since Qc > Ksp (1.0 x 10-6), precipitation will occur. b. Calculate Qc, the ion product of the solution, using the concentrations in the problem


Qc = [Pb2+][Cl-]2 Qc = (0.0035)(0.15)2 = 7.87 x 10-5 Since Qc > Ksp (1.6 x 10-5), precipitation will occur.

698 CHAPTER 18

18.40 a. Calculate Qc, the ion product of the solution, using the concentrations in the problem as the concentrations present after mixing and assuming no precipitation. Then, compare Qc with Ksp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of Ksp; then, use parentheses for the concentrations.

Qc = [Sr2+][CrO4

2-] Qc = (0.012)(0.0015) = 1.80 x 10-5 Since Qc < Ksp (3.5 x 10-5), no precipitation will occur. b. Calculate Qc, the ion product of the solution, using the concentrations in the problem


Qc = [Mg2+][CO3

2-] Qc = (0.0012)(0.041) = 8.61 x 10-5 Since Qc > Ksp (1.0 x 10-5), precipitation will occur. 18.41 Calculate the ion product, Qc, after preparation of the solution, assuming no

precipitation has occurred. Compare it with the Ksp. Qc = [Pb2+][CrO4

2-] Qc = (5.0 x 10-4)(5.0 x 10-5) = 2.50 x 10-8 M2 (> Ksp of 1.8 x 10-14) The solution is supersaturated before equilibrium is reached. At equilibrium,

precipitation occurs, and the solution is saturated. 18.42 Calculate the ion product, Qc, after preparation of the solution and assuming no

precipitation has occurred. Compare it with the Ksp. Qc = [Pb2+][SO4

2-] Qc = (5.0 x 10-4)(1.0 x 10-5) = 5.00 x 10-9 M2 (< Ksp of 1.7 x 10-8) At equilibrium, no precipitation occurs, and the solution is unsaturated.


18.43 Calculate the concentrations of Mg2+ and OH-, assuming no precipitation. Use a total volume of 1.0 L + 1.0 L, or 2.0 L. (Note the concentrations are halved when the volume is doubled.)

[Mg2+] =

0.0020 mol x 1.0 LL

2.0 L = 1.00 x 10-3 M

[OH-] =

0.00010 mol x 1.0 LL

2.0 L = 5.00 x 10-5 M


Qc = (Mg2+)(OH-)2 = (1.00 x 10-3)(5.00 x 10-5)2 = 2.50 x 10-12 Because Qc is less than the Ksp of 1.8 x 10-11, no precipitation occurs, and the solution

is unsaturated. 18.44 Calculate the concentrations of Ca2+ and SO4

2-, assuming no precipitation. Use a total volume of 0.045 L + 0.055 L, or 0.100 L.

[Ca2+] =

0.015 mol x 0.045 LL

0.100 L = 0.00675 M

[SO42-] =

0.010 mol x 0.055 LL

0.100 L = 0.00550 M


Qc = [Ca2+][SO4

2-] = (0.00675)(0.00550) = 3.71 x 10-5 Because Qc is greater than the Ksp of 2.4 x 10-5, the solution is supersaturated before

equilibrium is reached. At equilibrium, precipitation occurs, and the solution is saturated.

700 CHAPTER 18

18.45 Calculate the concentrations of Ba2+ and F-, assuming no precipitation. Use a total volume of 0.045 L + 0.075 L, or 0.120 L.

[Ba2+] =

0.0015 mol x 0.045 LL

0.120 L = 5.625 x 10-4 M

[F-] =

0.0025 mol x 0.075 LL

0.120 L = 1.56 x 10-3 M

Calculate the ion product and compare it to Ksp.

Qc = [Ba2+][F-]2 = (5.625 x 10-4)(1.56 x 10-3)2 = 1.36 x 10-9 Because Qc is less than the Ksp of 1.0 x 10-6, no precipitation occurs, and the solution

is unsaturated. 18.46 Calculate the concentrations of Pb2+ and Cl-, assuming no precipitation. Use a total

volume of 0.040 L + 0.065 L, or 0.105 L.

[Pb2+] =

0.010 mol x 0.065 LL

0.105 L = 6.19 x 10-3 M

[Cl-] =

0.035 mol x 0.040 LL

0.105 L = 1.33 x 10-2 M


Qc = [Pb2+][Cl-]2 = (6.19 x 10-3)(1.33 x 10-2)2 = 1.09 x 10-6 Because Qc is less than the Ksp of 1.6 x 10-5, no precipitation occurs, and the solution

is unsaturated.


18.47 A mixture of CaCl2 and K2SO4 can only precipitate CaSO4 because KCl is soluble. Use the Ksp expression to calculate the [Ca2+] needed to just begin precipitating the 0.020 M SO4

2- (in essentially a saturated solution). Then, convert to moles. [Ca2+][SO4

2-] = Ksp = 2.4 x 10-5

[Ca2+] = sp2-

4

K

[SO ] =

-5

-22.4 x 102.0 x 10

= 1.20 x 10-3 M

The number of moles in 1.5 L of this calcium-containing solution is (mol CaCl2 = mol Ca2+):

1.5 L x (1.20 x 10-3) mol/L = 1.80 x 10-3 = 0.0018 mol CaCl2

18.48 A mixture of MgSO4 and NaOH can only precipitate Mg(OH)2 because Na2SO4 is

soluble. Use the Ksp expression to calculate the [Mg2+] needed to just begin precipitating the 0.040 M OH- (in essentially a saturated solution). Then, convert to moles and finally to grams.

[Mg2+][OH-]2 = Ksp = 1.8 x 10-11

[Mg2+] = sp- 2

K

[OH ] =

-11

21.8 x 10(0.040)

= 1.125 x 10-8 M

The number of moles, and grams, in 0.456 L (456 mL) of this magnesium-containing

solution is (mol MgSO4 = mol Mg2+).

0.456 L x (1.125 x 10-8 mol/L) = 5.13 x 10-9 mol MgSO4

(5.13 x 10-9 mol MgSO4) x (120 g MgSO4/1 mol MgSO4)

= 6.156 x 10-7 = 6.2 x 10-7 g MgSO4 18.49 Because the AgNO3 solution is relatively concentrated, ignore the dilution of the

solution of Cl- and I- from the addition of AgNO3. The [Ag+] just as the AgCl begins to precipitate can be calculated from the Ksp expression for AgCl using [Cl-] = 0.015 M. Therefore, for AgCl

[Ag+][Cl-] = Ksp

(continued)

702 CHAPTER 18

[Ag+][0.015] = 1.8 x 10-10

[Ag+] = -101.8 x 10

0.015 = 1.20 x 10-8 M

The [I-] at this point can be obtained by substituting the [Ag+] into the Ksp expression for

AgI. Therefore, for AgI [Ag+][I-] = Ksp

[1.20 x 10-8][I-] = 8.3 x 10-17

[I-] = 8-

-17

10 x 1.210 x 8.3 = 6.91 x 10-9 = 6.9 x 10-9 M

18.50 Because the problem asks you to find [Cl-] at the point when Ag2CrO4 just begins to

precipitate, we start with the Ksp expression for Ag2CrO4, and ignore temporarily what may have happened to the Cl- ion. Also, because the AgNO3 solution is relatively concentrated, ignore the dilution of the solution of Cl- and CrO4

2- from the addition of AgNO3. The [Ag+] just as the Ag2CrO4 begins to precipitate can be calculated from the Ksp expression for Ag2CrO4 using [CrO4

2-] = 0.015 M. Thus, for Ag2CrO4 [Ag+]2[CrO4

2-] = Ksp [Ag+]2[0.015] = 1.1 x 10-12

[Ag+] = -121.1 x 10

0.015 = 8.56 x 10-6 M

The [Cl-] at this point can be obtained by substituting the [Ag+] into the Ksp expression

for AgCl. Therefore, for AgCl [Ag+][Cl-] = Ksp

[8.56 x 10-6][Cl-] = 1.8 x 10-10

[Cl-] = 6-

-10

10 x 8.5610 x 1.8 = 2.102 x 10-5 = 2.1 x 10-5 M

Because this Cl- concentration is extremely small compared to the initial 0.015 M concentration, you can also deduce that essentially all the chloride ion precipitates before the Ag2CrO4 begins to precipitate.


18.51 The net ionic equation is BaF2(s) + 2 H3O+(aq) Ba2+(aq) + 2 HF(aq) + 2 H2O(l) 18.52 The net ionic equation is PbCO3(s) + 2 H3O+(aq) Pb2+(aq) + CO2(g) + 2 H2O(l) 18.53 Calculate the value of K for the reaction of H3O+ with both the SO4

2- and F- anions as they form by the slight dissolving of the insoluble salts. These constants are the reciprocals of the Ka values of the conjugate acids of these anions.

F-(aq) + H3O+(aq) HF(aq) + H2O(l)

K = aK

1 = 4-10 x 6.8

1 = 1.47 x 103

SO4

2-(aq) + H3O+(aq) HSO4-(aq) + H2O(l)

K = a2K1 =

2-10 x 1.11 = 90.9

Because K for the fluoride ion is relatively larger, more BaF2 will dissolve in acid than BaSO4.

18.54 Calculate the value of K for the reaction of H3O+ with both the SO4

2- and PO43- anions

as they form by the slight dissolving of the insoluble salts. These constants are the reciprocals of the Ka values of the conjugate acids of these anions.

PO4

3-(aq) + H3O+(aq) HPO42-(aq) + H2O(l)

K = a3K1 =

13-10 x 4.81 = 2.08 x 1012

SO4

2-(aq) + H3O+(aq) HSO4-(aq) + H2O(l)

K = a2K1 =

2-10 x 1.11 = 90.9

Because K for the phosphate ion is relatively larger, more Ca3(PO4)2 will dissolve in acid than CaSO4.

704 CHAPTER 18

18.55 The equation is Cu+(aq) + 2 CN-(aq) Cu(CN)2

-(aq) The Kf expression is

Kf = 2-

-2

][CN ][Cu

]+

[Cu(CN) = 1.0 x 1016

18.56 The equation is Ni2+(aq) + 6 NH3(aq) Ni(NH3)6

2+(aq) The Kf expression is

Kf = 6

32

263

][NH ][Ni

])+

+[Ni(NH = 5.6 x 108

18.57 Assume the only [Ag+] is that in equilibrium with the Ag(CN)2

- formed from Ag+ and CN-. (In other words, assume all the 0.015 M Ag+ reacts with CN- to form 0.015 M Ag(CN)2

-.) Subtract the CN- that forms the 0.015 M Ag(CN)2

- from the initial 0.100 M CN-. Use this as the starting concentration of CN- for the usual concentration table.

[0.100 M NaCN - (2 x 0.015 M)] = 0.070 M starting CN-

Conc. (M) Ag(CN)2

- Ag+ + 2CN-

Starting 0.015 0 0.070

Change -x +x +2x


the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.015 and 2x is negligible compared to 0.070.

Kf = -

2+ -

Ag(CN) ][Ag ][CN ]2[ = 2

(0.015 - x)(x)(0.070 + 2x)

≅ 2(0.015)

(x)(0.070) ≅ 5.6 x 1018

Rearrange and solve for x = [Ag+]:

x ≅ (0.015) ÷ [(5.6 x 1018)(0.070)2] ≅ 5.46 x 10-19 = 5.5 x 10-19 M


18.58 You can make a rough assumption that most of the Zn(OH)42- that dissociates forms

Zn2+ and OH- ions. Assemble the usual concentration table, using 0.20 M as the starting concentration for Zn(OH)4

2- and assuming the starting concentrations of Zn2+ and OH- ions are zero.

Conc. (M) Zn(OH)4

2- Zn2+ + 4OH-

Starting 0.20 0 0

Change -x +x +4x

Equilibrium 0.20 - x x 4x Even though this reaction is the opposite of the equation for the formation constant,

the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.20.

Kf = 2-

42+ - 4(OH) ]

[Zn ][OH ][Zn = 4

(0.20 - x)(x)(4x)

≅ 4(0.20)

(x)(4x) ≅ 2.8 x 1015

Rearrange and solve for x = [Zn2+]:

x ≅ 5(0.20) = 1.

15 )10 x (2.8 x 25694 x 10-4 = 1.9 x 10-4 M

18.59 Start by calculating the [Cd2+] in equilibrium with the Cd(NH3)4

2+ formed from Cd2+ and NH3. Then, use the [Cd2+] to decide whether or not CdC2O4 will precipitate by calculating the ion product and comparing it with the Ksp of 1.5 x 10-8 for CdC2O4. Assume all the 0.0020 M Cd2+ reacts with NH3 to form 0.0020 M Cd(NH3)4

2+, and calculate the remaining NH3. Use these as the starting concentrations for the usual concentration table.

[0.10 M NH3 - (4 x 0.0020 M)] = 0.092 M starting NH3

Conc. (M) Cd(NH3)4

2+ Cd2+ + 4NH3

Starting 0.0020 0 0.092

Change -x +x +4x

Equilibrium 0.0020 - x x 0.092 + 4x

(continued)

706 CHAPTER 18

Even though this reaction is the opposite of the equation for the formation constant, the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.0020 and 4x is negligible compared to 0.092.

Kf = 2+

3 42+ 4

3

(NH ) ][Cd ][NH ][Cd = 4

(0.0020 - x)(x)(0.092 + 4x)

≅ 4(0.0020)

(x)(0.092) ≅ 1.0 x 107

Rearrange and solve for x: x ≅ (0.0020) ÷ [(1.0 x 107)(0.092)4] ≅ 2.79 x 10-6 M ≅ [Cd2+] Now, calculate the ion product for CdC2O4: Qc = [Cd2+][C2O4

2-] = (2.79 x 10-6)(0.010) = 2.79 x 10-8 Because Qc is greater than the Ksp of 1.5 x 10-8, the solution is supersaturated before

equilibrium. At equilibrium, a precipitate will form, and the solution will be saturated. 18.60 Start by calculating the [Ni2+] in equilibrium with the Ni(NH3)6

2+ formed from Ni2+ and NH3. Then, use the [Ni2+] to decide whether or not Ni(OH)2 will precipitate by calculating the ion product and comparing it with the Ksp of 2.0 x 10-15 for Ni(OH)2. Assume all the 0.0020 M Ni2+ reacts with NH3 to form 0.0020 M Ni(NH3)6

2+, and calculate the remaining NH3. Use these as starting concentrations for the usual concentration table.

[0.10 M NH3 - (6 x 0.0020 M)] = 0.088 M starting NH3

Conc. (M) Ni(NH3)62+ Ni2+ + 6NH3

Starting 0.0020 0 0.088

Change -x +x +6x


the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming that x is negligible compared to 0.0020 and 6x is negligible compared to 0.088.

(continued)


Kf = 2+

3 62+ 6

3

i(NH ) ][Ni ][NH ][N = 6

(0.0020 - x)(x)(0.088 + 6x)

≅ 6(0.0020)

(x)(0.088) ≅ 5.60 x 108

Rearrange and solve for x: x ≅ (0.0020) ÷ [5.6 x 108 x (0.088)6] ≅ 7.69 x 10-6 M ≅ [Ni2+] Now, calculate the ion product for Ni(OH)2: Qc = [Ni2+][OH-]2 = (7.69 x 10-6)(0.010)2 = 7.69 x 10-10 Because Qc is greater than the Ksp of 2.0 x 10-15, the solution is supersaturated before

equilibrium. At equilibrium, a precipitate will form, and the solution will be saturated. 18.61 Using the rule from Chapter 15, obtain the overall equilibrium constant for this reaction

from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation.

CdC2O4(s) Cd2+(aq) + C2O4

2-(aq) Ksp = 1.5 x 10-8

Cd2+(aq) + 4NH3(aq) Cd(NH3)42+(aq) Kf = 1.0 x 107

CdC2O4(s) + 4NH3(aq) Cd(NH3)42+(aq) + C2O4

2-(aq)

Kc = Ksp x Kf = 0.15 Assemble the usual table using 0.10 M as the starting concentration of NH3 and x as

the unknown concentration of Cd(NH3)42+ formed.

Conc. (M) CdC2O4(s) + 4NH3 Cd(NH3)4

2+ + C2O42-

Starting 0.10 0 0

Change -4x +x +x

Equilibrium 0.10 - 4x x x The equilibrium-constant expression can now be used. Simply substitute all the exact

equilibrium concentrations into the equilibrium-constant expression.

0.15 = )4x - 0.10(

)x( = ]NH[

]OC[])NH(Cd[ = K

4

2

43

-242

+243

c

(continued)

708 CHAPTER 18

Take the square root of both sides of the two right-hand terms, rearrange into a quadratic equation, and solve for x:

2x

(0.10 - 4x ) = 0.387

16x2 - 3.38x + 0.010 = 0

x = 23.38 (-3.38) - 4(16)(0.010)

2(16)±

= 0.208 (too large) and 3.001 x 10-3

Using the smaller root, x = 3.001 x 10-3 = 3.0 x 10-3 M (molar solubility of CdC2O4)

18.62 Using the rule from Chapter 15, obtain the overall equilibrium constant for this reaction

from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation.

NiS(s) Ni2+(aq) + S2-(aq) Ksp = 3 x 10-19

Ni2+(aq) + 6NH3(aq) Ni(NH3)62+(aq) Kf = 5.6 x 108

NiS(s) + 6NH3(aq) Ni(NH3)62+(aq) + S2-(aq)

Kc = Ksp x Kf = 1.7 x 10-10 Assemble the usual table using 0.10 M as the starting concentration of NH3 and x as

the unknown concentration of Ni(NH3)62+ formed.

Conc (M) NiS(s) + 6NH3 Ni(NH3)6

2+ + S2-

Starting 0.10 0 0

Change -6x +x +x


equilibrium concentrations into the equilibrium-constant expression. It will be necessary to simplify the exact equation to obtain a solution without using a higher-order equation.

10-6

2

63

-2+263

c 10 x 1.7 = )6x - 0.10(

)x( = ]NH[

]S[])NH(iN[ = K

(continued)


Take the square root of both sides of the two right-hand terms. Note that the resulting equation contains an x3 term and an x term.

3x

(0.10 - 6x) = 1.3 x 10-5

Assume 6x is negligible compared to 0.10 M, and solve the above equation for x:

3x

(0.10) = 1.3 x 10-5

x = 1.3 x 10-8 = 1 x 10-8 M (molar solubility of NiS in 0.10 M NH3)

18.63 The Pb2+, Cd2+, and Sr2+ ions can be separated in two steps: (1) Add HCl to precipitate

only the Pb2+ as PbCl2, leaving the others in solution. (2) After pouring the solution away from the precipitate, add 0.3 M HCl and H2S to precipitate only the CdS away from the Sr2+ ion, whose sulfide is soluble under these conditions.

18.64 The Hg2+, Ca2+, and Na+ ions can be separated in two steps: (1) Add 0.3 M HCl and

H2S to precipitate only the HgS, leaving the others in solution. (2) After pouring the solution away from the precipitate, add NH3 and (NH4)2HPO4 to precipitate only the Ca3(PO4)2 away from the Na+ ion, whose phosphate is soluble under these conditions.

18.65 a. Ag+ is not possible because no precipitate formed with HCl. b. Ca2+ is not possible if the compound contains only one cation because Mn2+ is

indicated by the evidence. However, if the compound consists of two or more cations, Ca2+ is possible because no reactions were described involving Ca2+.

c. Mn2+ is possible because a precipitate was obtained with basic sulfide ion. d. Cd2+ is not possible because no precipitate was obtained with acidic sulfide solution. 18.66 PbCrO4 is not the compound formed from H2S because it doesn't contain sulfide ion.

CdS is possible because it contains sulfide ion and precipitates in acidic sulfide (Analytical Group II). Neither MnS nor Ag2S is possible because neither falls into Analytical Group II.

710 CHAPTER 18

Solutions to General Problems

18.67 Assemble the usual concentration table. Let x equal the molar solubility of PbSO4. When x mol PbSO4 dissolves in one L of solution, x mol Pb2+ and x mol SO4

2- form.

Conc. (M) PbSO4(s) Pb2+ + SO42-

Starting 0 0

Change +x +x


solve for x. [Pb2+][SO4

2-] = Ksp

(x)(x) = x2 = 1.7 x 10-8

x = )10 x 1.7( 8- = 1.303 x 10-4 = 1.3 x 10-4 M 18.68 Assemble the usual concentration table. Let x equal the molar solubility of HgS. When x

mol HgS dissolves in one L of solution, x mol Hg2+ and x mol S2- form.

Conc. (M) HgS(s) Hg2+ + S2-

Starting 0 0

Change +x +x


solve for x. [Hg2+][S2-] = Ksp

(x)(x) = x2 = 1.6 x 10-52

x = )10 x 1.6( 52- = 1.26 x 10-26 = 1.3 x 10-26 M


18.69 Let x equal the molar solubility of Hg2Cl2. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression.

Conc. (M) Hg2Cl2(s) Hg2

2+ + 2Cl-

Starting 0 0

Change +x +2x

Equilibrium x 2x [Hg2

2+][Cl-]2 = Ksp (x)(2x)2 = 4x3 = 1.3 x 10-18

x = -18

3 1.3 x 104

= 6.87 x 10-7 M

a. Molar solubility = 6.87 x 10-7 = 6.9 x 10-7 M

b. L

mol 10 x -76.87 x 2 2472.1 g Hg Cl1 mol

= 3.24 x 10-4 = 3.2 x 10-4 g/L

18.70 Let x equal the molar solubility of MgNH4PO4. Assemble the usual concentration table,

and substitute from the table into the equilibrium-constant expression.

Conc. (M) MgNH4PO4(s) Mg2+ + NH4+ + PO4

3-

Starting 0 0 0

Change +x +x +x

Equilibrium x x x [Mg2+][NH4

+][PO43-] = K sp

(x)(x)(x) = x3 = 2.5 x 10-13

x = 3 -132.5 x 10 = 6.299 x 10-5 M


b. -59 x 10 mol

L 6.29 x 4 4137.3 g MgNH PO

1 mol

= 8.648 x 10-3 = 8.6 x 10-3 g/L

712 CHAPTER 18

18.71 Let x equal the molar solubility of Ce(OH)3. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression.

Conc. (M) Ce(OH)3(s) Ce3+ + 3OH-

Starting 0 0

Change +x +3x

Equilibrium x 3x [Ce3+][OH-]3 = Ksp (x)(3x)3 = 27x4 = 2.0 x 10-20

x = -20

4 2.0 x 1027

= 5.216 x 10-6 M


b. [OH-] = 3x = 1.56 x 10-5 M pOH = 4.8054 = 4.81 18.72 Let x equal the molar solubility of Cu2Fe(CN)6. Assemble the usual concentration table,

and substitute from the table into the equilibrium-constant expression.

Conc. (M) Cu2Fe(CN)6 (s) 2Cu2+ + Fe(CN)64-

Starting 0 0

Change +2x +x

Equilibrium 2x x [Cu2+]2[Fe(CN)6

4-] = Ksp (2x)2(x) = 4x3 = 1.3 x 10-16

x = -16

3 1.3 x 104

= 3.19 x 10-6 M


b. -6.19 x 10 mol

L 3 x 2339.0 g Cu Fe(CN)

1 mol

6 = 1.08 x 10-3 = 1.1 x 10-3 g/L


18.73 Calculate the pOH from the pH, and then convert pOH to [OH-]. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression.

pOH = 14.00 - 8.80 = 5.20 [OH-] = antilog (-pOH) = antilog (-5.20) = 6.309 x 10-6 M


Starting 0 0

Change +x —

Equilibrium x 6.309 x 10-6 Ksp = [Mg2+][OH-]2 = (x)(6.309 x 10-6)2 = 1.8 x 10-11

x = 26-

-11

)10 x (6.30910 x 1.8 = 0.452 = 0.45 M

L

mol 0.452 x mol 1Mg(OH) g 58.3 2 = 26.3 = 26 g/L

18.74 Calculate the pOH from the pH, and then convert pOH to [OH-]. Then, assemble the

usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. Let 2x equal the molar solubility of the Ag+ ion because you are seeking x, the molar solubility of Ag2O.

pOH = 14.00 - 10.50 = 3.50 [OH-] = antilog (-pOH) = antilog (-3.50) = 3.16 x 10-4 M

Conc. (M) Ag2O(s) + 2H2O 2Ag+ + 2OH-

Starting 0 0

Change x +2x —

Equilibrium 2x 3.16 x 10-4 Kc = [Ag+]2[OH-]2 = (2x)2(3.16 x 10-4)2 = 2.0 x 10-8

(continued)

714 CHAPTER 18

x = 24-

8-

)10 x (3.16 x 410 x 2.0 = 0.224 = 0.22 M

L

mol 0.224 x mol 1

O Agg 231.7 2 = 51.9 = 52 g/L

18.75 Let x equal the change in M of Mg2+ and 0.10 M equal the starting OH- concentration.

Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. Assume 2x is negligible compared to 0.10 M, and perform an approximate calculation.


Starting 0 0.10

Change +x +2x

Equilibrium x 0.10 + 2x Ksp = [Mg2+][OH-]2 = (x)(0.10 + 2x)2 ≅ (x)(0.10)2 ≅ 1.8 x 10-11

x = 2

-11

(0.10)10 x 1.8 = 1.80 x 10-9 = 1.8 x 10-9 M

18.76 Let x equal the change in M of Al3+ and 0.0010 M equal the starting OH- concentration.

Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. Assume 3x is negligible compared to 0.0010 M, and perform an approximate calculation.

Conc. (M) Al(OH)3(s) Al3+ + 3OH-

Starting 0 0.0010

Change +x +3x

Equilibrium x 0.0010 + 3x Ksp = [Al3+][OH-]3 = (x)(0.0010 + 3x)3 ≅ (x)(0.0010)3 ≅ 4.6 x 10-33

x = 3

-33

(0.0010)10 x 4.6 = 4.60 x 10-24 = 4.6 x 10-24 M


18.77 To begin precipitation, you must add just slightly more sulfate ion than that required to give a saturated solution. Use the Ksp expression to calculate the [SO4

2-] needed to just begin precipitating the 0.0030 M Ca2+.

[Ca2+][SO4

2-] = Ksp = 2.4 x 10-5

[SO42-] = sp

2+

K

[Ca ] =

-5

-32.4 x 103.0 x 10

= 8.00 x 10-3 M

When the sulfate ion concentration slightly exceeds 8.0 x 10-3 M, precipitation begins.

18.78 To begin precipitation, you must add just slightly more chromate ion than that required

to give a saturated solution. Use the Ksp expression to calculate the [CrO42-] needed to

just begin precipitating the 0.0025 M Sr2+. [Sr2+][CrO4

2-] = Ksp = 3.5 x 10-5

[CrO42-] = sp

2+

K

[Sr ] =

-5

-33.5 x 102.5 x 10

= 1.40 x 10-2 M

When the chromate ion concentration slightly exceeds 1.4 x 10-2 M, precipitation begins.

18.79 Calculate the concentrations of Pb2+ and Cl-. Use a total volume of 3.20 L + 0.80 L, or

4.00 L.

[Pb2+] =

-31.25 x 10 mol x 3.20 LL

4.00 L = 1.00 x 10-3 M

[Cl-] =

-15.0 x 10 mol x 0.80 LL

4.00 L = 1.00 x 10-1 M


Qc = [Pb2+][Cl-]2 = (1.00 x 10-3)(1.00 x 10-1)2 = 1.00 x 10-5 M3 Because the Qc is less than the Ksp of 1.6 x 10-5, no precipitation occurs, and the

solution is not saturated.

716 CHAPTER 18

18.80 Rounded answer: Qc equals 1.8 x 10-11 (equals Ksp, so solution is saturated). Calculate the concentrations of Mg2+ and OH-. Use a total volume of 0.150 L + 0.050 L, or 0.200 L.

[Mg2+] =

-52.4 x 10 mol x 0.150 LL

0.200 L = 1.80 x 10-5 M

[OH-] =

-34.0 x 10 mol x 0.050 LL

0.200 L = 1.00 x 10-3 M


Qc = [Mg2+][OH-]2 = (1.80 x 10-5)(1.00 x 10-3)2 = 1.80 x 10-11 Because Qc equals the Ksp of 1.8 x 10-11, no precipitation occurs, and the solution is

saturated. 18.81 A mixture of AgNO3 and NaCl can precipitate only AgCl because NaNO3 is soluble.

Use the Ksp expression to calculate the [Cl-] needed to prepare a saturated solution (just before precipitating the 0.0015 M Ag+). Then, convert to moles and finally to grams.

[Ag+][Cl-] = Ksp = 1.8 x 10-10

[Cl-] = sp+

K

[Ag ] =

-10

-31.8 x 101.5 x 10

= 1.20 x 10-7 M

The number of moles and grams in 0.785 L (785 mL) of this chloride-containing

solution is

0.785 L x (1.20 x 10-7 mol/L) = 9.42 x 10-8 mol Cl- = 9.42 x 10-8 mol NaCl

(9.42 x 10-7 mol NaCl) x (58.5 g NaCl/1mol) = 5.51 x 10-6 = 5.5 x 10-6 g NaCl This amount of NaCl is too small to weigh on a balance.


18.82 A mixture of BaCl2 and Na2SO4 can precipitate only BaSO4 because NaCl is soluble. Use the Ksp expression to calculate the [SO4

2-] needed to prepare a saturated solution (just before precipitating the 0.0028 M Ba2+). Then, convert to moles and finally to grams.

[Ba2+][SO4

2-] = Ksp = 1.1 x 10-10

[SO42-] = sp

2+

K

[Ba ] =

-10

-31.1 x 102.8 x 10

= 3.93 x 10-8 M

The number of moles and grams in 0.435 L of this sulfate-containing solution is

(0.435 L x 3.93 x 10-8 mol/L) = 1.709 x 10-8 mol SO42- = 1.709 x 10-8 mol Na2SO4

(1.709 x 10-8 mol Na2SO4) x (142 g/1 mol Na2SO4)

= 2.42 x 10-6 = 2.4 x 10-6 g Na2SO4 18.83 From the magnitude of Kf, assume Fe3+ and SCN- react essentially completely to form

2.00 M Fe(SCN)2+ at equilibrium. Use 2.00 M as the starting concentration of Fe(SCN)2+ for the usual concentration table.

Conc. (M) Fe3+ + SCN- FeSCN2+

Starting 0 0 2.00

Change +x +x -x

Equilibrium x x 2.00 - x Substitute all the exact equilibrium concentrations into the formation-constant

expression; then, simplify the exact equation by assuming x is negligible compared to 2.00 M.

22-+3

+2

f 10 x 9.0 x

)2.00( )x()x(

) x- 2.00( = ]SCN[]Fe[

]FeSCN[ = K ≅≅


x = 210 x 9.0

2.00 = 4.71 x 10-2 = 4.7 x 10-2 M

Fraction dissociated = [(4.71 x 10-2) ÷ (2.00)] x 100%

= 0.0235 (< 0.03, so acceptable)

718 CHAPTER 18

18.84 From the magnitude of Kf, assume Co2+ and SCN- react essentially completely to form 2.00 M Co(SCN)+ at equilibrium. Use 2.00 M as the starting concentration of Co(SCN)+ for the usual concentration table.

Conc. (M) Co2+ + SCN- CoSCN+

Starting 0 0 2.00

Change +x +x -x

Equilibrium x x 2.00 - x Substitute all the exact equilibrium concentrations into the formation-constant

expression; then, simplify the exact equation by assuming x is negligible compared to 2.00 M.

22-+2

+

f 10 x 1.0 x

)2.00( )x()x(

) x- 2.00( = ]SCN[]Co[

]CoSCN[ = K ≅≅


x = 210 x 1.0

2.00 = 0.141 = 0.14 M

The value of x = 0.14 affects the term 2.00 - x slightly in the second significant figure.

From the quadratic formula, you obtain x = 0.137, which also rounds to 0.14. 18.85 Obtain the overall equilibrium constant for this reaction from the product of the

individual equilibrium constants of the two individual equations whose sum gives this equation:

AgBr(s) Ag+(aq) + Br-(aq) Ksp = 5.0 x 10-13

Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) Kf = 1.7 x 107

AgBr(s) + 2NH3(aq) Ag(NH3)22+(aq) + Br-(aq)

Kc = Ksp x Kf = 8.50 x 10-6 Assemble the usual table using 5.0 M as the starting concentration of NH3 and x as the

unknown concentration of Ag(NH3)2+ formed.

(continued)


Conc. (M) AgBr(s) + 2NH3 Ag(NH3)2+ + Br-

Starting 5.0 0 0

Change -2x +x +x


equilibrium concentrations into the equilibrium-constant expression; it will not be necessary to simplify the equation because taking the square root of both sides removes the x2 term.

6-2

2

23

-+23

c 10 x 058. = )2x - 5.0(

x = ]NH[

]rB[])(NHAg[ = K

Take the square root of both sides of the two right-hand terms and solve for x.

3-10 x 1592. = )2x - 5.0(

x

x + (5.8 x 10-3) x = 1.4575 x 10-2

x = 1.449 x 10-2 = 1.4 x 10-2 M (the molar solubility of AgBr in 5.0 M NH3)

18.86 Obtain the overall equilibrium constant for this reaction from the product of the

individual equilibrium constants of the two individual equations whose sum gives this equation:

AgI(s) Ag+(aq) + I-(aq) Ksp = 8.3 x 10-17


AgI(s) + 2NH3(aq) Ag(NH3)2+(aq) + I-(aq)

Kc = Ksp x Kf = 1.41 x 10-9 Assemble the usual table using 2.0 M as the starting concentration of NH3 and x as the

unknown concentration of Ag(NH3)2+ formed.

(continued)

720 CHAPTER 18

Conc. (M) AgI(s) + 2NH3 Ag(NH3)2+ + I-

Starting 2.0 0 0

Change -2x +x +x


equilibrium concentrations into the equilibrium-constant expression; it will not be necessary to simplify the equation because taking the square root of both sides removes the x2 term.

9-2

2

23

-+23

c 10 x 141. = )2x - 2.0(

x = ]NH[

]I[])(NHAg[ = K

Take the square root of both sides of the two right-hand terms and solve for x.

5-10 x 573. = )2x - 2.0(

x

x + (7.5 x 10-5)x = 7.50 x 10-5

x = 7.499 x 10-5 = 7.5 x 10-5 M (the molar solubility of AgI in 2.0 M NH3)

18.87 Start by recognizing that, because each zinc oxalate produces one oxalate ion, the

solubility of zinc oxalate equals the oxalate concentration: ZnC2O4(s) + 4NH3 Zn(NH3)4

2+ + C2O42-

Thus, the [C2O4

2-] = 3.6 x 10-4 M. Now, calculate the [Zn2+] in equilibrium with the oxalate ion using the Ksp expression for zinc oxalate.

Ksp = [Zn2+][C2O4

2-] = 1.5 x 10-9

[Zn2+] = 4-

-9

10 x 3.610 x 1.5 = 4.16 x 10-6 = 4.2 x 10-6 M

(continued)


To calculate Kf, the [Zn(NH3)42+] term must be calculated. This can be done by

recognizing that the molar solubility of ZnC2O4 is the sum of the concentration of Zn2+ and Zn(NH3)4

2+ ions:

Molar solubility of ZnC2O4 = [Zn2+] + [Zn(NH3)42+]

3.6 x 10-4 = (4.16 x 10-6) + [Zn(NH3)4

2+] [Zn(NH3)4

2+] = (3.6 x 10-4) - (4.16 x 10-6) = 3.558 x 10-4 M

Now, the [NH3] term must be calculated by subtracting the ammonia in [Zn(NH3)42+]

from the starting NH3 of 0.0150 M:

[NH3] = 0.0150 - 4 [Zn(NH3)4

2+] = 0.0150 - 4(3.558 x 10-4) = 0.01357 M

Solve for Kf by substituting the known concentrations into the Kf expression:

Kf = 2+

3 42+ 4

3

n(NH ) ][Zn ][NH ][Z =

-4

-6 43.558 x 10

(4.16 x 10 )(0.01357) = 2.52 x 109 = 2.5 x 109

18.88 Start by recognizing that, because each cadmium oxalate produces one oxalate ion,

the solubility of cadmium oxalate equals the oxalate concentration: CdC2O4(s) + 4NH3 Cd(NH3)4

2+ + C2O42-

Thus, the [C2O4

2-] = 6.1 x 10-3 M. Now, calculate the [Cd2+] in equilibrium with the oxalate ion using the Ksp expression for cadmium oxalate.

Ksp = [Cd2+][C2O4

2-] = 1.5 x 10-8

[Cd2+] = 3-

-8

10 x 6.110 x 1.5 = 2.46 x 10-6 = 2.5 x 10-6 M

To calculate Kf, the [Cd(NH3)4

2+] term must be calculated. This can be done by recognizing that the molar solubility of CdC2O4 is the sum of the concentration of Cd2+ and Cd(NH3)4

2+ ions:

Molar solubility of CdC2O4 = [Cd2+] + [Cd(NH3)42+]

6.1 x 10-3 = (2.46 x 10-6) + [Cd(NH3)4

2+] (continued)

722 CHAPTER 18

[Cd(NH3)42+] = (6.1 x 10-3) - (2.46 x 10-6) ≅ 6.1 x 10-3 M

Now the [NH3] term must be calculated by subtracting the ammonia in [Cd(NH3)4

2+] from the starting NH3 of 0.0150 M:

[NH3] = 0.150 - 4[Cd(NH3)4

2+] = 0.150 - 4(6.1 x 10-3) = 0.1256 M Solve for Kf by substituting the known concentrations into the Kf expression:

Kf = 2+

3 42+ 4

3

(NH ) ][Cd ][NH ][Cd =

-3

-6 46.1 x 10

(2.46 x 10 )(0.1256) = 9.96 x 106 = 1.0 x 107

18.89 The OH- formed by ionization of NH3 (to NH4

+ and OH- ) is a common ion that will precipitate Mg2+ as the slightly soluble Mg(OH)2 salt. The simplest way to treat the problem is to calculate the [OH- ] of 0.10 M NH3 before the soluble Mg2+ salt is added. As the soluble Mg2+ salt is added, the [Mg2+] will increase until the solution is saturated with respect to Mg(OH)2 (the next Mg2+ ions to be added will precipitate). Calculate the [Mg2+] at the point at which precipitation begins.

To calculate the [OH-] of 0.10 M NH3, let x equal the mol/L of NH3 that ionize, forming

x mol/L of NH4+ and x mol/L of OH- and leaving (0.10 - x) M NH3 in solution. We can

summarize the situation in tabular form:

Conc. (M) NH3 + H2O NH4+ + OH-

Starting 0.10 ~0 0

Change -x +x +x

Equilibrium 0.10 - x x x The equilibrium-constant equation is:

)0.10(

x ) x- 0.10(

x = ]NH[

]OH[]NH[ = K22

3

-+4

c ≅

The value of x can be obtained by rearranging and taking the square root:

M 10 x 431. = 0.10 x 10 x 1.8 ]OH[ 3-5-- ≅ (continued)


Note that (0.10 - x) is not significantly different from 0.10, so x can be ignored in the (0.10 - x) term.

Now, use the Ksp of Mg(OH)2 to calculate the [Mg2+] of a saturated solution of Mg(OH)2, which essentially will be the Mg2+ ion concentration when Mg(OH)2 begins to precipitate.

[Mg2+] = 23-

-11

2sp

)10 x 1.34(10 x 1.8 =

]OH[

K−

= 1.00 x 10-5 = 1.0 x 10-5 M

18.90 The OH- formed by ionization of N2H4 (to N2H5

+ and OH-) is a common ion that will precipitate Mg2+ as the slightly soluble Mg(OH)2 salt. The simplest way to treat the problem is to calculate the [OH-] of 0.20 M N2H4 before the soluble Mg2+ salt is added. As the soluble Mg2+ salt is added, the [Mg2+] will increase until the solution is saturated with respect to Mg(OH)2 (the next Mg2+ ions to be added will precipitate). Calculate the [Mg2+] at the point at which precipitation begins.

To calculate the [OH-] of 0.20 M N2H4, let x equal the mol/L of N2H4 that ionizes, forming x mol/L of N2H5 and x mol/L of OH- and leaving (0.20 - x) M N2H4 in solution. We can summarize the situation in tabular form:

Conc. (M) N2H4 + H2O N2H5

+ + OH-

Starting 0.20 ~0 0

Change -x +x +x

Equilibrium 0.20 - x x x The equilibrium-constant equation is:

)0.20(

x ) x- 0.20(

x = ]HN[

]OH[]HN[ = K

22

42

-+52

c ≅

The value of x can be obtained by rearranging and taking the square root:

M 10 x 385. = 0.20 x 10 x 1.7 ]OH[ 4-6-- ≅ (continued)

724 CHAPTER 18

Note that (0.20 - x) is not significantly different from 0.10, so x can be ignored in the (0.20 - x) term.

Now, use the Ksp of Mg(OH)2 to calculate the [Mg2+] of a saturated solution of Mg(OH)2, which essentially will be the Mg2+ ion concentration when Mg(OH)2 begins to precipitate.

[Mg2+] = 24-

-11

2sp

)10 x 5.83(10 x 1.8 =

]OH[

K−

= 5.29 x 10-5 = 5.3 x 10-5 M

18.91 a. Use the solubility information to calculate Ksp. The reaction is Cu(IO3)2(s) Cu2+(aq) + 2IO3

-(aq)

2.7 x 10-3 2 x (2.7 x 10-3) Ksp = [Cu2+][IO3

-]2 = [2.7 x 10-3][2 x (2.7 x 10-3)]2 = 7.87 x 10-8 Set up an equilibrium. The reaction is Cu(IO3)2(s) Cu2+(aq) + 2IO3

-(aq)

y 0.35 + 2y ≈ 0.35 Ksp = [y][0.35]2 = 7.87 x 10-8

Molar solubility = y = 6.42 x 10-7 = 6.4 x 10-7 M b. Set up an equilibrium. The reaction is Cu(IO3)2(s) Cu2+(aq) + 2IO3

-(aq)

y + 0.35 ≈ 0.35 2y Ksp = [0.35][2y]2 = 7.87 x 10-8

Molar solubility = y = 2.37 x 10-4 = 2.4 x 10-4 M c. Yes, Cu(IO3)2 is a 1:2 electrolyte. It takes two IO3

- ions to combine with one Cu2+ ion. The IO3

- ion is involved as a square term in the Ksp expression.


18.92 a. Use the solubility information to calculate Ksp. The reaction is Pb(IO3)2(s) Pb2+(aq) + 2IO3

-(aq)

4.0 x 10-5 M 2 x (4.0 x 10-5) M Ksp = [Pb2+][IO3

-]2 = [4.0 x 10-5][2 x (4.0 x 10-5)]2 = 2.56 x 10-13 Set up an equilibrium. The reaction is Pb(IO3)2(s) Pb2+(aq) + 2IO3

-(aq)

y + 0.15 ≈ 0.15 2y Ksp = [0.15][2y]2 = 2.56 x 10-13

Molar solubility = y = 6.53 x 10-7 = 6.5 x 10-7 M b. Because of the common ion effect or a consideration of Le Chatelier’s principle, the

molar solubility would be less. 18.93 a. Set up an equilibrium. The reaction and equilibrium-constant expression are PbI2(s) Pb2+(aq) + 2I-(aq) Ksp = [Pb2+][I-]2 = 6.5 x 10-9 The Pb2+ ion concentration is 0.0150 M. Plug this in, and solve for the iodine ion

concentration.

M10 x 6.6 = 10 x 856. = 0.0150

10 x 6.5 = ]I[ 4-4-9-

-

b. Solve the equilibrium-constant expression for the lead ion concentration.

23-

-9+2

)10 x 2.0(10 x 6.5 = ]Pb[ = 1.63 x 10-3 M

The percent of the lead(II) ion remaining in solution is

Percent Pb2+ remaining = -3x 10

0.0150 1.63 x 100% = 10.9 = 11 percent

726 CHAPTER 18

18.94 a. Set up an equilibrium. The reaction and equilibrium-constant expression are CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = [Ca2+][F-]2 = 3.4 x 10-11 Solve the equilibrium-constant expression for the fluoride ion concentration. The

calcium ion concentration is 0.00750 M.

[F- ] = 0.00750

10 x 3.4 11-

= 6.73 x 10-5 = 6.7 x 10-5 M

b. Solve the equilibrium-constant expression for the calcium ion concentration. The

fluoride ion concentration is 9.5 x 10-4 M.

24-

-11+2

)10 x 9.5(10 x 3.4 = ]Ca[ = 3.77 x 10-5 M

The percent of the calcium ion precipitated is

Percent Ca2+ precipitated = -50.00750 - 3.77 x 10

0.00750 x 100% = 99.4

= 99 percent 18.95 a. The reaction and equilibrium-constant expression are Co(OH)2(s) Co2+(aq) + 2OH-(aq) Ksp = [Co2+][OH-]2

From the molar solubility, [Co2+] = 5.4 x 10-6 M and [OH-] = 2 x (5.4 x 10-6) M. Therefore,

Ksp = [5.4 x 10-6][2 x (5.4 x 10-6)]2 = 6.30 x 10-16 = 6.3 x 10-16

b. From the pOH (14 - pH), the [OH-] = 10-3.57 = 2.69 x 10-4 M. The molar solubility is equal to the cobalt ion concentration at equilibrium.

[Co2+] = 24-

-16

2-sp

)10 x 2.69(10 x 6.3 =

]OH[

K = 8.71 x 10-9 = 8.7 x 10-9 M

c. The common ion effect (OH-) in part (b) decreases the solubility of Co2+.


18.96 a. The reaction and equilibrium-constant expression are Be(OH)2(s) Be2+(aq) + 2OH-(aq) Ksp = [Be2+][OH-]2

From the molar solubility, [Be2+] = 8.6 x 10-7 M and [OH-] = 2 x (8.6 x 10-7) M. Therefore,

Ksp = [8.6 x 10-7][2 x (8.6 x 10-7)]2 = 2.54 x 10-18 = 2.5 x 10-18 b. This is a buffer system. The reaction is NH3 + H2O NH4

+ + OH- Kb = 1.8 x 10-5

Set up the equilibrium, and calculate the OH- ion concentration.

Kb = 5--

3

-+4 10 x 1.8 =

)1.50(]OH[)0.25( =

]NH[]OH[]NH[

[OH-] = 1.08 x 10-4 M The molar solubility is equal to the beryllium ion concentration at equilibrium.

[Be2+] = 24-

-18

2-sp

)10 x 1.08(10 x 2.5 =

]OH[

K = 2.14 x 10-10 = 2.1 x 10-10 M

c. The common ion effect (a Le Chatelier stress) decreases the solubility of Be2+.

18.97 a. AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 1.8 x 10-10


AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl-(aq)

K = Ksp x Kf = (1.8 x 10-10)(1.7 x 107) = 3.06 x 10-3 = 3.1 x 10-3

(continued)

728 CHAPTER 18

b. The equilibrium-constant expression is

3-2

2

23

-+23 10 x 3.06 =

)0.80(y =

]NH[

]lC[])NH(Ag[

y = [Ag(NH3)2

+] = 0.0443 = 0.044 M

mol AgCl dissolved = mol Ag(NH3)2+ = 0.044 mol/L x 1.00 L = 0.044 mol

mol NH3 reacted = 0.0443 mol Ag(NH3)2+ x 3

+3 2

2 mol NH1 mol Ag(NH )

= 0.0886 mol

The total moles of NH3 added = 0.80 M x 1.00 L + 0.0886 = 0.889 = 0.89 mol

18.98 a. AgBr(s) Ag+(aq) + Br-(aq) Ksp = 5.0 x 10-13

Ag+(aq) + 2S2O32-(aq) Ag(S2O3)2

3- Kf = 2.9 x 1013

AgBr(s) + 2S2O32-(aq) Ag(S2O3)2

3-(aq) + Br-(aq)

K = Ksp x Kf = (5.0 x 10-13)(2.9 x 1013) = 14.5 = 15

mol AgBr = 2.5 g AgBr x 1 mol AgBr187.77 g AgBr

= 0.0133 mol

mol Ag(S2O3)2

3- = mol AgBr = 0.0133 mol

5.14 = ]OS[

)0133.0( = ]OS[

]rB[])OS(Ag[2-2

32

2

2-232

--2232

Take the square root of both sides to obtain the moles of S2O3

2- at equilibrium.

14.5 = ]OS[

0.0133-2

32

[S2O3

2-] = 0.00349 mol/L

(continued)


The moles of S2O32- in Ag(S2O3)2

3- = 2 x 0.0133 = 0.0266 mol

Total moles of S2O32- = 0.0266 + 0.00349 = 0.0301 = 0.030 mol

The moles of Na2S2O3 added = moles of S2O3

2- = 0.030 mol 18.99 a. The moles of NH4Cl (molar mass 53.49 g/mol) are given by

mol NH4Cl = 26.7 g x 4

4

1 mol NH Cl53.49 g NH Cl

= 0.4992 mol

The reaction and equilibrium-constant expression are

NH3 + H2O NH4+ + OH- Kb =

]NH[]OH[]NH[

3

-+4 = 1.8 x 10-5

Solve the equilibrium-constant expression for [OH-]. The molarity of the NH4Cl is 0.4992 mol/1.0 L = 0.4992 M.

)0.4992(

)(4.2)10 x 1.8( = ]OH[-5

- = 1.51 x 10-4 = 1.5 x 10-4 M

b. The reaction and equilibrium-constant expression are Mg(OH)2 Mg2+ + 2OH- Ksp = [Mg2+][OH-]2 = 1.8 x 10-11

Solving for [Mg2+] gives

24-

-11+2

)10 x 1.51()10 x 1.8( = ]Mg[ = 7.89 x 10-4 = 7.9 x 10-4 M

The percent Mg2+ that has been removed is given by

Percent Mg2+ removed = -40.075 - 7.89 x 10

0.075 x 100% = 98.9 = 99 percent

730 CHAPTER 18

18.100 a. The moles of (NH4)2SO4 (molar mass 132.15 g/mol) are given by

mol (NH4)2SO4 = 75.8 g x 4 2 4

4 2 4

1 mol (NH ) SO132.15 g (NH ) SO

= 0.5736 mol

The moles of NH4+ = 2 x 0.5736 = 1.147 mol. The reaction and equilibrium-

constant expression are

NH3 + H2O NH4+ + OH- Kb =

+ -4

3

[NH ][OH ][NH ]

= 1.8 x 10-5

Solve the equilibrium-constant expression for [OH-]. The molarity of the NH4Cl is 1.147 mol/1.0 L = 1.147 M.

)1.147(

)(1.6)10 x 1.8( = ]OH[-5

- = 2.51 x 10-5 = 2.5 x 10-5 M

b. The reaction and equilibrium-constant expression are Mn(OH)2 Mn2+ + 2OH- Ksp = [Mn2+][OH-]2 = 4.6 x 10-14

Solving for [Mn2+] gives

25-

-14+2

)10 x 2.51()10 x 4.6( = ]Mn[ = 7.30 x 10-5 = 7.3 x 10-5 M

The percent Mn2+ that has been removed is given by

Percent Mn2+ removed = -50.058 - 7.30 x 10

0.058 x 100% = 99.9 percent

Nearly 100 percent of the Mn2+ has been removed.

Solutions to Cumulative-Skills Problems

18.101 Using the Ksp of 1.1 x 10-21 for ZnS, calculate the [S2-] needed to maintain a saturated solution (without precipitation):

[S2-] = +24-

-21

Zn M 10 x 1.510 x 1.1 = 7.33 x 10-18 M

(continued)


Next, use the overall H2S ionization expression to calculate the [H3O+] needed to achieve this [S2-] level:

[H3O+] = M 10 x 7.33

M) (0.10 10 x 1.118-

20- = 1.225 x 10-2 M

Finally, calculate the buffer ratio of [SO4

2-]/[HSO4-] from the H2SO4 Ka2 expression

where Ka2 has the value 1.1 x 10-2:

][HSO

][SO-

4

2-4 =

]O[H

K

3

a2+

= 2-

-2

10 x 1.22510 x 1.1 =

1.0000.8979

If [HSO4

-] = 0.20 M, then [SO4

2-] = 0.8979 x 0.20 M = 0.1795 = 0.18 M 18.102 Using the Ksp of 4 x 10-21 for CoS, calculate the [S2-] needed to maintain a saturated

solution (without precipitation):

[S2-] = +24-

-21

Co M 10 x 1.810 x 4 = 2.22 x 10-17 M

Next, use the overall H2S ionization expression to calculate the [H3O+] needed to

achieve this [S2-] level:

[H3O+] = M 10 x 2.22

M) (0.10 10 x 1.117-

20- = 7.04 x 10-3 M

Finally, calculate the buffer ratio of [SO4

2-]/[HSO4-] from the H2SO4 Ka2 expression

where Ka2 has the value 1.1 x 10-2:

][HSO

][SO-

4

2-4 =

]O[H

K

3

a2+

= 3-

-2

10 x 7.0410 x 1.1 =

1.0001.56

If [HSO4

-] = 0.20 M, then [SO4

2-] = 1.56 x 0.20 M = 0.312 = 0.3 M

732 CHAPTER 18

18.103 Begin by solving for [H3O+] in the buffer. Ignoring changes in [HCHO2] as a result of ionization in the buffer, you obtain

[H3O+] ≅ 1.7 x 10-4 x M 0.20M 0.45 = 3.825 x 10-4 M

You should verify that this approximation is valid (you obtain the same result from the

Henderson-Hasselbalch equation). The equilibrium for the dissolution of CaF2 in acidic solution is obtained by subtracting twice the acid ionization of HF from the solubility equilibrium of CaF2:

CaF2(s) Ca2+(aq) + 2F-(aq) Ksp

2H3O+(aq) + 2F-(aq) 2HF(aq) + 2 H2O(l) 1/(Ka)2

2H3O+(aq) + CaF2(s) Ca2+(aq) + 2HF(aq) + 2 H2O(l) Kc = Ksp/(Ka)2 Therefore, Kc = (3.4 x 10-11) ÷ (6.8 x 10-4)2 = 7.35 x 10-5. In order to solve the

equilibrium-constant equation, you require the concentration of HF, which you obtain from the acid-ionization constant for HF.

]HF[

]F[]OH[ = K

-+3

a

6.8 x 10-4 = 3.825 x 10-4 x ]HF[]F[ -

]HF[]F[ -

= 1.778, or [F-] = 1.778 [HF]

Let x be the solubility of CaF2 in the buffer. Then, [Ca2+] = x and [F-] + [HF] = 2x.

Substituting from the previous equation, you obtain

2x = 1.778[HF] + [HF] = 2.778[HF], or [HF] = 2x/2.778

You can now substitute for [H3O+] and [HF] into the equation for Kc.

5-24-

2

2+3

2+2

c 10 x 7.35 = )10 x 3.825(

)2.778/2x(x = ]OH[

]HF[]Ca[ = K

7.35 x 10-5 = (3.543 x 106)x3

x3 = 2.075 x 10-11

x = 2.748 x 10-4 = 2.7 x 10-4 M


18.104 Begin by solving for [H3O+] in the buffer. Ignoring changes in [HC2H3O2] as a result of ionization in the buffer, you obtain

[H3O+] ≅ 1.7 x 10-5 x M 0.20M 0.45 = 3.825 x 10-5 M

You should verify that this approximation is valid (you obtain the same result from the

Henderson-Hasselbalch equation). The equilibrium for the dissolution of MgF2 in acidic solution is obtained by subtracting twice the acid ionization of HF from the solubility equilibrium of MgF2:

MgF2(s) Mg2+(aq) + 2F-(aq) Ksp

2H3O+(aq) + 2F-(aq) 2HF(aq) + 2 H2O(l) 1/(Ka)2

2H3O+(aq) + MgF2(s) Mg2+(aq) + 2HF(aq) + 2 H2O(l) Kc = Ksp/(Ka)2 Therefore, Kc = (6.5 x 10-9) ÷ (6.8 x 10-4)2 = 1.406 x 10-2. In order to solve the

equilibrium-constant equation, you require the concentration of HF, which you obtain from the acid-ionization constant for HF.

]HF[

]F[]OH[ = K

-+3

a

6.8 x 10-4 = 3.825 x 10-5 x ]HF[]F[ -

]HF[]F[ -

= 17.78, or [F-] = 17.78 [HF]

Let x be the solubility of MgF2 in the buffer. Then, [Mg2+] = x and [F-] + [HF] = 2x.

Substituting from the previous equation, you obtain

2x = 17.78[HF] + [HF] = 18.78[HF], or [HF] = 2x/18.78

You can now substitute for [H3O+] and [HF] into the equation for Kc.

Kc = 25-

2

2+3

2+2

)10 x 3.825(

)78.18/2x(x = ]OH[

]HF[]Mg[ = 1.406 x 10-2

1.406 x 10-2 = (7.752 x 106)x3

x3 = 1.812 x 10-9

x = 1.219 x 10-3 = 1.2 x 10-3 M

734 CHAPTER 18

18.105 The net ionic equation is Ba2+(aq) + 2OH-(aq) + Mg2+(aq) + SO4

2-(aq) BaSO4(s) + Mg(OH)2(s) Start by calculating the mol/L of each ion after mixing and before precipitation. Use a

total volume of 0.0450 + 0.0670 L = 0.112 L.

M of SO42- and Mg2+ = (0.350 mol/L x 0.0670 L) ÷ 0.112 L = 0.2094 M

M of Ba2+ = (0.250 mol/L x 0.0450 L) ÷ 0.112 L = 0.1004 M

M of OH- = (2 x 0.250 mol/L x 0.0450 L) ÷ 0.112 L = 0.20089 M

Assemble a table showing the precipitation of BaSO4 and Mg(OH)2.

Conc. (M) Ba2+ + SO42- + Mg2+ + 2OH-

BaSO4(s) + Mg(OH)2(s)

Starting 0.1004 0.2094 0.2094 0.20089

Change -0.1004 -0.1004 -0.1004 -0.20089

Equilibrium 0.0000 0.1090 0.1090 0.0000

To calculate [Ba2+], use the Ksp expression for BaSO4.

[Ba2+] = 2-

4

-10

SO M 090.10

10 x 1.1 = 1.009 x 10-9 = 1.0 x 10-9 M

[SO4

2-] = 0.1090 = 0.109 M Calculate the [OH-] using the Ksp expression for Mg(OH)2:

[OH-] = +2

11-

Mg M 0.109010 x 1.8 = 1.28 x 10-5 = 1.3 x 10-5 M

[Mg2+] = 0.1090 = 0.109 M


18.106 The net ionic equation is Pb2+(aq) + 2Cl-(aq) + 2Ag+(aq) + SO4

2-(aq) PbSO4(s) + 2AgCl(s) Start by calculating the mol/L of each ion after mixing and before precipitation. Use a

total volume of 0.0250 + 0.0500 L = 0.0750 L.

M of SO42- = (0.0150 mol/L x 0.0500 L) ÷ 0.0750 L = 0.01000 M

M of Pb2+ = (0.0100 mol/L x 0.0250 L) ÷ 0.0750 L = 0.003333 M

M of Ag+ = (2 x 0.0150 mol/L x 0.0500 L) ÷ 0.0750 L = 0.02000 M

M of Cl- = (2 x 0.0100 mol/L x 0.0250 L) ÷ 0.07500 = 0.006666 M

Assemble a table showing the precipitation of PbSO4 and AgCl.

Conc. (M) Pb2+ + SO42- + Ag+ + Cl-

PbSO4(s) + AgCl(s)

Starting 0.003333 0.01000 0.02000 0.006667

Change -0.003333 -0.003333 -0.006667 -0.006667

Equilibrium 0.000000 0.006667 0.013333 0.000000

To calculate [Pb2+], use the Ksp expression for PbSO4.

[Pb2+] = 2-

4

-8

SO M 0.006667

10 x 1.7 = 2.549 x 10-6 = 2.50 x 10-6 M

[SO4

2-] = 0.006667 = 0.0067M Calculate the [Cl-] using the Ksp expression for AgCl:

[Cl-] = + AgM 0.01333

10 x 1.8 -10 = 1.3503 x 10-8 = 1.4 x 10-8 M

[Ag+] = 0.01333 = 0.0133 M

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